A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10,000 N/m. The spring compresses by

Solution & Explanation

1. Formulate the Work-Energy Theorem

According to the Work-Energy Theorem, the net work done by all forces acting on a body is equal to the change in its kinetic energy:

$$W_{\text{net}} = \Delta K$$

$$W_{\text{spring}} + W_{\text{friction}} = K_{\text{final}} - K_{\text{initial}}$$

Let $$x$$ be the maximum compression of the spring in meters ($$\text{m}$$) at the instant the block is brought completely to rest. Let us define the physical parameters from the problem statement:

• Mass of the block: $$m = 2 \,\, \text{kg}$$
• Initial velocity: $$v = 4 \,\, \text{m/s}$$
• Kinetic friction force: $$f_k = 15 \,\, \text{N}$$
• Spring constant: $$k = 10,000 \,\, \text{N/m}$$
• Final velocity: $$v_f = 0 \,\, \text{m/s}$$ (since it comes to a stop)

2. Set Up the Quadratic Energy Balance Equation

We substitute the mathematical expressions for each work component into the energy conservation layout:

• Work done by the restoring spring force: $$W_{\text{spring}} = -\frac{1}{2} \cdot k \cdot x^2$$
• Work done by the retarding friction force over displacement $$x$$: $$W_{\text{friction}} = -f_k \cdot x$$

Substituting these into the relation yields:

$$-\frac{1}{2} \cdot k \cdot x^2 - f_k \cdot x = 0 - \frac{1}{2} \cdot m \cdot v^2$$

Multiply the entire equation by $$-1$$ to make the coefficients positive:

$$\frac{1}{2} \cdot k \cdot x^2 + f_k \cdot x = \frac{1}{2} \cdot m \cdot v^2$$

Now, plug in the numerical values into the equation:

$$\frac{1}{2} \cdot (10,000) \cdot x^2 + 15 \cdot x = \frac{1}{2} \cdot (2) \cdot (4)^2$$

$$5000 \cdot x^2 + 15 \cdot x = 16$$

$$5000 \cdot x^2 + 15 \cdot x - 16 = 0$$

3. Solve the Quadratic Equation for Compression ($$x$$)

Using the general quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ to find the roots of our equation:

$$x = \frac{-15 \pm \sqrt{(15)^2 - 4 \cdot (5000) \cdot (-16)}}{2 \cdot 5000}$$

$$x = \frac{-15 \pm \sqrt{225 + 320,000}}{10,000}$$

$$x = \frac{-15 \pm \sqrt{320,225}}{10,000}$$

Since $$\sqrt{320,225} \approx 565.88$$, and discarding the negative root as compression length must be positive:

$$x = \frac{-15 + 565.88}{10,000}$$

$$x = \frac{550.88}{10,000} \approx 0.0551 \,\, \text{m}$$

4. Convert the Final Answer to Centimeters

To convert the calculated SI unit value from meters to centimeters ($$\text{cm}$$), we multiply the result by $$100$$:

$$x = 0.0551 \cdot 100 \,\, \text{cm} \approx 5.5 \,\, \text{cm}$$

Concept Check: The initial kinetic energy of the block ($$16 \,\, \text{J}$$) is split between storing elastic potential energy inside the spring and overcoming the resistive drag of friction. Because the spring constant is exceptionally stiff ($$10,000 \,\, \text{N/m}$$), the total stopping distance is restricted to a small displacement of just $$5.5 \,\, \text{cm}$$.

Correct Option Key: Option A ($$5.5 \,\, \text{cm}$$)